Use the order of operations to simplify the expression below. 990 ÷ (12-9)
1 answer:
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Answer:
a.
b.
c.
No, it is not unusual if at least 1 lives up to 3.
Step-by-step explanation:
Given
Represent the probability that a 2 year old snake will live to 3 with P(Live);
Solving (a): Probability that two selected will live to 3 years.
Both snakes have a chance of 0.98861 to live up to 3 years.
So, the required probability is:
<em>--- Approximated</em>
Solving (b): Probability that seven selected will live to 3 years.
All 7 snakes have a chance of 0.98861 to live up to 3 years.
So, the required probability is:
Where
<em>--- Approximated</em>
Solving (c): Probability that at least one of seven selected will not live to 3 years.
In probabilities, the following relationship exist:
So, first we need to calculate the probability that none of the 7 lived up to 3.
If the probability that one lived up to 3 years is 0.98861, then the probability than one do not live up to 3 years is 1 - 0.98861
This gives:
The probability that none of the 7 lives up to 3 is:
Substitute this value for P(None) in
---- Approximated
No, it is not unusual if at least 1 lives up to 3.
This is so because the above results, which is 1 shows that it is very likely for at least one of the seven to live up to 3 years